How do I get the path to the current script with Node.js?

How would I get the path to the script in Node.js?

I know there's process.cwd, but that only refers to the directory where the script was called, not of the script itself. For instance, say I'm in /home/kyle/ and I run the following command:

node /home/kyle/some/dir/file.js

If I call process.cwd(), I get /home/kyle/, not /home/kyle/some/dir/. Is there a way to get that directory?

I found it after looking through the documentation again. What I was looking for were the __filename and __dirname module-level variables.

• __filename is the file name of the current module. This is the resolved absolute path of the current module file. (ex:/home/kyle/some/dir/file.js)
• __dirname is the directory name of the current module. (ex:/home/kyle/some/dir)

So basically you can do this:

fs.readFile(path.resolve(__dirname, 'settings.json'), 'UTF-8', callback);

Use resolve() instead of concatenating with '/' or '' else you will run into cross-platform issues.

Note: __dirname is the local path of the module or included script. If you are writing a plugin which needs to know the path of the main script it is:

require.main.filename

or, to just get the folder name:

require('path').dirname(require.main.filename)

Use __dirname!!

__dirname

The directory name of the current module. This the same as the path.dirname() of the __filename.

Example: running node example.js from /Users/mjr

console.log(__dirname);
// Prints: /Users/mjr
console.log(path.dirname(__filename));
// Prints: /Users/mjr

https://nodejs.org/api/modules.html#modules_dirname

For ESModules you would want to use: import.meta.url

This command returns the current directory:

var currentPath = process.cwd();

For example, to use the path to read the file:

var fs = require('fs');
fs.readFile(process.cwd() + "\\text.txt", function(err, data)
{
    if(err)
        console.log(err)
    else
        console.log(data.toString());
});

Node.js 10 supports ECMAScript modules, where __dirname and __filename are no longer available.

Then to get the path to the current ES module one has to use:

import { fileURLToPath } from 'url';

const __filename = fileURLToPath(import.meta.url);

And for the directory containing the current module:

import { dirname } from 'path';
import { fileURLToPath } from 'url';

const __dirname = dirname(fileURLToPath(import.meta.url));

When it comes to the main script it's as simple as:

process.argv[1]

From the Node.js documentation:

> ## process.argv > > An array containing the command line arguments. The first element will be 'node', the second element will be the path to the JavaScript file. The next elements will be any additional command line arguments.

If you need to know the path of a module file then use __filename.

var settings = 
    JSON.parse(
        require('fs').readFileSync(
            require('path').resolve(
                __dirname, 
                'settings.json'),
            'utf8'));

Every Node.js program has some global variables in its environment, which represents some information about your process and one of it is __dirname.

I know this is pretty old, and the original question I was responding to is marked as duplicate and directed here, but I ran into an issue trying to get jasmine-reporters to work and didn't like the idea that I had to downgrade in order for it to work. I found out that jasmine-reporters wasn't resolving the savePath correctly and was actually putting the reports folder output in jasmine-reporters directory instead of the root directory of where I ran gulp. In order to make this work correctly I ended up using process.env.INIT_CWD to get the initial Current Working Directory which should be the directory where you ran gulp. Hope this helps someone.

var reporters = require('jasmine-reporters');
var junitReporter = new reporters.JUnitXmlReporter({
  savePath: process.env.INIT_CWD + '/report/e2e/',
  consolidateAll: true,
  captureStdout: true
 });

You can use process.env.PWD to get the current app folder path.

If you are using pkg to package your app, you'll find useful this expression:

appDirectory = require('path').dirname(process.pkg ? process.execPath : (require.main ? require.main.filename : process.argv[0]));

• process.pkg tells if the app has been packaged by pkg.

• process.execPath holds the full path of the executable, which is /usr/bin/node or similar for direct invocations of scripts (node test.js), or the packaged app.

• require.main.filename holds the full path of the main script, but it's empty when Node runs in interactive mode.

• __dirname holds the full path of the current script, so I'm not using it (although it may be what OP asks; then better use appDirectory = process.pkg ? require('path').dirname(process.execPath) : (__dirname || require('path').dirname(process.argv[0])); noting that in interactive mode __dirname is empty.

• For interactive mode, use either process.argv[0] to get the path to the Node executable or process.cwd() to get the current directory.

Use the basename method of the path module:

var path = require('path');
var filename = path.basename(__filename);
console.log(filename);

Here is the documentation the above example is taken from.

As Dan pointed out, Node is working on ECMAScript modules with the "--experimental-modules" flag. Node 12 still supports __dirname and __filename as above.

---

If you are using the --experimental-modules flag, there is an alternative approach.

The alternative is to get the path to the current ES module:

const __filename = new URL(import.meta.url).pathname;

And for the directory containing the current module:

import path from 'path';

const __dirname = path.dirname(new URL(import.meta.url).pathname);

NodeJS exposes a global variable called __dirname.

__dirname returns the full path of the folder where the JavaScript file resides.

So, as an example, for Windows, if we create a script file with the following line:

console.log(__dirname);

And run that script using:

node ./innerFolder1/innerFolder2/innerFolder3/index.js

The output will be: C:\Users...<project-directory>\innerFolder1\innerFolder2\innerFolder3

index.js within any folder containing modules to export

const entries = {};
for (const aFile of require('fs').readdirSync(__dirname, { withFileTypes: true }).filter(ent => ent.isFile() && ent.name !== 'index.js')) {
  const [ name, suffix ] = aFile.name.split('.');
  entries[name] = require(`./${aFile.name}`);
}

module.exports = entries;

This will find all files in the root of the current directory, require and export every file present with the same export name as the filename stem.

If you want something more like $0 in a shell script, try this:

var path = require('path');

var command = getCurrentScriptPath();

console.log(`Usage: ${command} <foo> <bar>`);

function getCurrentScriptPath () {
    // Relative path from current working directory to the location of this script
    var pathToScript = path.relative(process.cwd(), __filename);

    // Check if current working dir is the same as the script
    if (process.cwd() === __dirname) {
        // E.g. "./foobar.js"
        return '.' + path.sep + pathToScript;
    } else {
        // E.g. "foo/bar/baz.js"
        return pathToScript;
    }
}